Question

If A forms FCC unit cell and B occupies 50% octahedral void & 50% tetrahedral void then the formula of the compound is​

Answer 1

Answer:

AB is the formula

hope the answer will help you

Answer 2

Answer:

The empirical formula of the compound containing atoms A and B observed is $A_{4} B_{6}$ or $A_{2}B_{3}$.

Explanation:

Given data,

Atom A forms unit cell = FCC ( face-centred cubic ) unit cell.

Atom B occupies - $50\%$ octahedral voids and $50\%$ tetrahedral voids.

The compound's empirical formula containing atoms A and B =?

As we know,

• In FCC, the number of atoms is $4$.

Therefore,

• The number of atoms A = $4$

Also, in FCC,

• The number of octahedral voids = $4$
• The number of tetrahedral voids = $8$

Therefore,

• The $50\%$ of octahedral voids = $2$
• The $50\%$ of tetrahedral voids = $4$

Thus,

• The number of atoms B = $2+4$ = $6$

Hence, the compound's empirical formula is $A_{4} B_{6}$ or $A_{2}B_{3}$.